functions

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\(\large \color{black}{\begin{align} &\normalsize \text{Find the maximim value of the function}\hspace{.33em}\\~\\ & \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\ \end{align}}\)
interval given ?
with out interval this maximum value diverges to infinity.

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this question means that i have to minimize \(\large \color{black}{\begin{align} x^2-3x+2 \hspace{.33em}\\~\\ \end{align}}\) right ?
should be, or els... i mean there is not maximum value here, if you are taking over \((-\infty,~+\infty)\)
wait
1/that ? or, just that ?
lol its \(\large \color{black}{\begin{align} \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\ \end{align}}\)
\(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-3x+2} }\) and you want to find absolute minimum, right?
well, I can tell you that the limit as x approaches \(\pm\)infinity is going to be 0.
we can plot, I am guessing it is something like |dw:1435010239497:dw|
i want to find the maximum value of this \(\large \color{black}{\begin{align} \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\ \end{align}}\)
none
it is +infinity
but how ?
https://www.desmos.com/calculator
how i prove that with algebra
even the minimum doesn't exist.
with algebra, even with no calc ?
idk calculus
ic
lets c
\(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-3x+2} }\) it has vertical asymptotes, can you find them for me?
\(\large \color{black}{\begin{align} x=2,\ 1 \hspace{.33em}\\~\\ \end{align}}\)
yes
and, the best thing I can see here is to use the idea of a limit to see where range tends to the close we get to x=1 and x=2. I don't really know a purely algebraic proof for why this function has no max or min.
that has a max actually
yes?
https://www.desmos.com/calculator keep scrollin up
on a certain interval, no doubt (unless it has a domain gap)
i think it attains one max does not perhaps required any domain to get it or i think
your graph does not show up see here https://www.desmos.com/calculator
i don't think there is a max. that is due to an asymptote - two of them, there.
we can find the limit from the right and from the left as x approaches 1 and 2.
http://prntscr.com/7k4q1l here see that bottom part of the graph
with calculus we can prove it has a max value but @mathmath333 is looking for a different way
this graph goes further down and up if you scroll on desmos, but we can go ahead and take any limit that is closed from either side of any of the asymptotes you wish to choose.....
i mean here local max not abolute of course
how to find that local max
oh, but the question is asking for absolute doesn't it?
that is what i would think of course
first let's write that as 1/(x-1)(x-2)
i had thought of absolute too, just not sure about it!
yes the question is asking absolute, i just asked that of curiosity
this function is pretty weird. I wonder what would it model
i mean in real world
you want to then prove there is no max algebraically
yes ^
would that include using ideas from calc but not its techniques?
I am stubborn on the idea of a limit. A limit would simply show that as it approaches __ it can be infinitely large/small
ok i will try to bear with calculus
\(\large\color{black}{ \displaystyle \frac{1 }{x^2-3x+2} }\) we need to do only 1 asymptote for this
\(\large\color{black}{ \displaystyle \frac{1 }{x^2-3x+2} }\) lets plug in values that are less than, but close to 1
\(\large\color{black}{ \displaystyle f(0.9)=? }\)
= about 9
f(0.99)=99
try plugging it yourself for close values to 1
these values are called approaching "from the left" because they are a buit less than 1
now, try to do f(1.5) f(1.1) f(1.01) f(1.001) and you will see where it goes "from the right" - for values that are greater than 1
i got to go
sorry
suppose there exist a max max(1/(x^2-3x+2)=c 1/x^2-3x+21/c=l x^2-3x+2-l>0
just some random stuff lol
coming back..
ok i need to go please comment if u found something
let's first consider the domain \(D_f: \{ x\in \mathbb{R}: x\ne1, x\ne 2 \}\)
that is (-oo, 1)u(1, 2)u(2,oo)
we look at the sign of y when x<1 if y does not change the sign in that interval then i must not have a max seems that 1/(x-2)(x-1)>0 for x<1 same reasoning for x>2 ===> y>0 so only btw 1 and 2 left
from the graph we knew for 1

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